(a) Let us call the formula $\phi$ . Then, $\phi$ is assigned $T$ iff both terms are assigned $T$ iff $A=T$ implies $B=T$ and $A=F$ implies $C=T$ . This is the conditional ternary connective. Hence, here are the values for $A$ , $B$ and $C$ for which $\phi$ is true:

$$\begin{eqnarray*} A & B & C\\ T & T & T\\ T & T & F\\ F & F & T\\ F & T & T \end{eqnarray*}$$

If an implicant $\alpha$ of $\phi$ consists of some literals, then $\phi$ must be true when each literal is true, regardless of the values of other variables. And a prime implicant is such that there is no proper sub-implicant. In our case, a single literal would be an implicant iff there would be four same values ($T$ or $F$ ) in some column (corresponding to four possible values of the other variables), which is not the case. Now, columns $A$ and $B$ have a submatrix $2\times2$ such that values in each column are the same (the first two rows). These two rows correspond to all the possible values of the other variable $C$ . Since the values are $T$ and $T$ in the two columns of the submatrix, $A\wedge B$ is a prime implicant. Similarly, $\neg A\wedge C$ is another prime implicant (the last two rows of the columns $A$ and $C$ ). Finally, $B\wedge C$ is also a prime implicant (the first and last rows of the columns $B$ and $C$ ). Now, other implicants would necessary consist of at least three literals, corresponding to each single row of the matrix, but since we have already found an implicant corresponding to each row, those implicants would not be prime. Overall, the prime implicants of $\phi$ are $A\wedge B$ , $\neg A\wedge C$ and $B\wedge C$ .

(b) Those that cover all rows of the matrix, namely, $(A\wedge B)\vee(\neg A\wedge C)$ , and the disjunction of all three.